Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $y = \dfrac{a^2 - a}{8a - 40} \div \dfrac{a^2 + 3a - 4}{a - 5} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{a^2 - a}{8a - 40} \times \dfrac{a - 5}{a^2 + 3a - 4} $ First factor the quadratic. $y = \dfrac{a^2 - a}{8a - 40} \times \dfrac{a - 5}{(a - 1)(a + 4)} $ Then factor out any other terms. $y = \dfrac{a(a - 1)}{8(a - 5)} \times \dfrac{a - 5}{(a - 1)(a + 4)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ a(a - 1) \times (a - 5) } { 8(a - 5) \times (a - 1)(a + 4) } $ $y = \dfrac{ a(a - 1)(a - 5)}{ 8(a - 5)(a - 1)(a + 4)} $ Notice that $(a - 5)$ and $(a - 1)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ a\cancel{(a - 1)}(a - 5)}{ 8(a - 5)\cancel{(a - 1)}(a + 4)} $ We are dividing by $a - 1$ , so $a - 1 \neq 0$ Therefore, $a \neq 1$ $y = \dfrac{ a\cancel{(a - 1)}\cancel{(a - 5)}}{ 8\cancel{(a - 5)}\cancel{(a - 1)}(a + 4)} $ We are dividing by $a - 5$ , so $a - 5 \neq 0$ Therefore, $a \neq 5$ $y = \dfrac{a}{8(a + 4)} ; \space a \neq 1 ; \space a \neq 5 $